Signed double Roman k-domination in graphs

Signed double Roman k-domination in graphs

J. Amjadi

Department of Mathematics

Azarbaijan Shahid Madani University, Tabriz I.R. Iran

j-amjadi@azaruniv.ac.ir

Hong Yang

School of Information Science and Engineering Chengdu University, Chengdu 610106

China

yanghong01@cdu.edu.cn

S. Nazari-Moghaddam S.M. Sheikholeslami

Department of Mathematics

Azarbaijan Shahid Madani University, Tabriz I.R. Iran

s.nazari@azaruniv.ac.ir s.m.sheikholeslami@azaruniv.ac.ir

Zehui Shao

Institute of Computing Science and Technology Guangzhou University, Guangzhou 510006, China

zshao@gzhu.edu.cn Abstract

Let G = (V, E) be a simple and finite graph with vertex set V (G), and let k ≥ 1 be an integer. A signed double Roman k-dominating function (SDRkDF) on a graph G is a function f : V (G) → {−1, 1, 2, 3} such that (i) every vertex v with f (v) =−1 is adjacent to at least two vertices assigned with 2 or to at least one vertex w with f (w) = 3, (ii) every vertex v with f (v) = 1 is adjacent to at least one vertex w with f (w) ≥ 2 and (iii) 

u∈N [v]f (u)≥ k holds for any vertex v. The weight of an SDRkDF f is

u∈V (G)f (u), and the minimum weight of an SDRkDF is the signed double Roman k-domination number γ_sdR^k (G) of G. In this paper, we initiate the study of the signed double Roman k-domination number in graphs and we present lower and upper bounds for γ_sdR^k (T ). In addition we determine this parameter for some classes of graphs.

1 Introduction

Throughout this paper, G denotes a simple graph, with vertex set V = V (G) and edge set E = E(G). The order |V | of G is denoted by n = n(G). Denote by Kn

the complete graph and by C_n the cycle of order n. For every vertex v ∈ V , the open neighborhood of v is the set N (v) = {u ∈ V (G) | uv ∈ E(G)} and the closed neighborhood of v is the set N [v] = N (v)∪ {v}. The degree of a vertex v ∈ V is d(v) = |N(v)|. The minimum degree and the maximum degree of a graph G are denoted by δ = δ(G) and Δ = Δ(G), respectively. If every vertex of G has degree r, then G is said to be r-regular. The open neighborhood of a set S ⊆ V is the set N (S) = ∪v∈SN (v), and the closed neighborhood of S is the set N [S] = N (S)∪ S.

The distance d_G(u, v) between two vertices u and v in a connected graph G is the length of a shortest u−v path in G. The diameter of a graph G, denoted by diam(G), is the greatest distance between two vertices of G. The complement of a graph G is denoted by G. A leaf of G is a vertex with degree one and a support vertex is a vertex adjacent to a leaf. A tree T is a double star if it contains exactly two vertices that are not leaves. A double star with respectively p and q leaves attached at each support vertex is denoted by DS_p,q. A bipartite graph is one whose vertex set can be partitioned into two subsets X and Y , so that each edge has one end in X and one end in Y ; such a partition (X, Y ) is called a bipartition of the graph. A complete bipartite graph is a simple bipartite graph with bipartition (X, Y ) in which each vertex of X is joined to each vertex of Y ; if|X| = m and |Y | = n, such a graph is denoted by K_m,n.

A set S ⊆ V in a graph G is a dominating set if every vertex of G is either in S or adjacent to a vertex of S. The domination number γ(G) equals the minimum cardinality of a dominating set in G. For a comprehensive treatment of domination in graphs, see the monographs by Haynes, Hedetniemi, and Slater [9, 10].

For a subset S ⊆ V (G) of vertices of a graph G and a function f : V (G) → R, we define f (S) = 

x∈Sf (x). For a vertex v, we denoted f (N [v]) by f [v] for notional convenience.

A double Roman dominating function(DRDF) is a function f : V (G)→ {0, 1, 2, 3}

having the property that if f (v) = 0, then vertex v must have at least two neighbors assigned 2 under f or one neighbor with f (w) = 3, and if f (v) = 1, then vertex v must have at least one neighbor with f (w)≥ 2. The weight of a double Roman dom- inating function f is ω(f ) =

v∈V (G)f (v). The double Roman dominating number of G is the minimum weight of a double Roman dominating function on G. The double Roman domination was introduced by Beeler et al. [7] and has been studied by several authors [1, 2, 4, 6, 16, 17, 21].

A signed Roman k-dominating function (SRkDF) on a graph G is a function f : V → {−1, 1, 2} satisfying the conditions that (i) 

x∈N [v]f (x) ≥ k for each vertex v ∈ V , and (ii) every vertex u for which f(u) = −1 is adjacent to at least one vertex v for which f (v) = 2. The weight of an SRkDF is the sum of its function values over all vertices. The signed Roman k-domination number of G, denoted γ_sR^k (G), is the minimum weight of an SRkDF in G. The signed Roman k-domination number was introduced by Henning and Volkman in [11] and has been studied in

[12, 13, 18, 19, 20]. The special case k = 1 was introduced and investigated in [5]

and has been studied in [14, 15].

In this paper, we continue the study of the double Roman dominating functions on graphs. Inspired by the previous research on the signed Roman k-dominating function [11, 12], we define the signed double Roman k-dominating function as fol- lows.

Let k ≥ 1 be an integer. A function f : V (G) → {−1, 1, 2, 3} is a signed double Roman k-dominating function (SDRkDF) on a graph G if the following conditions are fulfilled:

(i) 

x∈N [v]f (v)≥ k for every vertex v ∈ V (G);

(ii) if f (v) =−1, then vertex v must have at least two neighbors with label 2 or one neighbor with label 3;

(iii) if f (v) = 1, then vertex v must have at least one neighbor with f (w)≥ 2.

The weight of an SDRkDF is the sum of its function values over all vertices. The signed double Roman k-domination number of G, denoted γ^k_sdR(G), is the minimum weight of an SDRkDF in G. A signed double Roman k-dominating function of G of weight γ_sdR^k (G) is called a γ_sdR^k (G)-function or γ_sdR^k -function of G. The special case k = 1 has been studied by Ahangar et al. [3]. If f is a signed double Roman k-dominating function of G and v ∈ V (G), then by definition we must have k ≤



x∈N [v]f (v) ≤

x∈N [v]3 = 3(deg(v) + 1) yielding deg(v)≥ k/3 − 1 and so δ(G) ≥ k/3−1. As the assumption δ(G) ≥ k/3−1 is necessary, we always assume that when we discuss γ_sdR^k (G), all graphs involved satisfy δ(G) ≥ k/3 − 1 and thus n(G) ≥ k/3.

An SDRkDF f can be represented by the ordered quadruple (V₋₁, V₁, V₂, V₃) of V (G) where V_i = {v ∈ V (G) | f(v) = i} for i ∈ {−1, 1, 2, 3}. In this paper we initiate the study of signed double Roman k-domination numbers in graphs and investigate their basic properties. In particular, we establish some sharp bounds on signed double Roman k-domination. In addition, we determine the signed double Roman k-domination number of some classes of graphs.

We make use of the following results in this paper.

Observation 1.1. If f = (V₋₁, V₁, V₂, V₃) is an SDRkDF on a graph G of order n, then

(a) |V₋₁| + |V₁| + |V₂| + |V₃| = n.

(b) ω(f ) =|V1| + 2|V2| + 3|V3| − |V−1|.

Observation 1.2. If k ≥ 1 is an integer and G is a graph of order n with δ(G) ≥

^k−1₂ , then γ_sdR^k (G)≤ 2n.

Proof. Clearly, the function f : V (G) → {−1, 1, 2, 3} defined by f(x) = 2 for x ∈ V (G) is an SDRkDF on G of weight 2n and thus γ_sdR^k (G)≤ 2n.

Let k≥ 2 and n ≥ ^k₂ be integers and let V be a set of size n. Define f_n^k : V → {−1, 1, 2, 3} as follows. If n + k ≡ 0 (mod 3), then let f_n^k assign 2 to ^n+k₃ elements of V and −1 to the remaining elements, if n + k ≡ 1 (mod 3), then let f_n^k assign 3 to one element of V , 2 to ^n+k−4₃ elements of V and −1 to the remaining elements, and if n + k ≡ 2 (mod 3), then let f_n^k assign 3 to two elements of V , 2 to ^n+k−8₃ elements of V and −1 to the remaining elements.

Observation 1.3. Let k ≥ 2 and n ≥ ^k₂ be integers. If n + k ≥ 6, then γ_sdR^k (K_n) = k.

Proof. For any vertex v ∈ V (Kn), we have γ_sdR^k (G) = f [v] ≥ k. Now we show that γ_sdR^k (K_n) ≤ k. If k is an even number and n = k/2, then the result follows from Observation 1.2. If k is an odd number or n≥ ^k₂ + 1, then it is easy to verify that the function f_n^k defined above, is an SDRkDF on K_nof weight k, and so γ_sdR^k (G)≤ k.

Thus γ_sdR^k (G) = k.

2 Basic properties and bounds

In this section we present basic properties of the signed double Roman k-dominating function.

Proposition 2.1. Let f = (V₋₁, V₁, V₂, V₃) be an SDRkDF on a graph G of order n.

If δ ≥ k − 1, then

(i) (3Δ + 3− k)|V3| + (2Δ + 2 − k)|V2| + (Δ + 1 − k)|V1| ≥ (δ + k + 1)|V−1|.

(ii) (3Δ + δ + 4)|V₃| + (2Δ + δ + 3)|V₂| + (Δ + δ + 2)|V₁| ≥ (δ + k + 1)n.

(iii) (Δ + δ + 2)ω(f ) ≥ (δ − Δ + 2k)n + (δ − Δ)|V₂| + 2(δ − Δ)|V₃|.

(iv) ω(f )≥ (δ − 3Δ + 2k − 2)n/(3Δ + δ + 4) + |V2| + 2|V3|.

Proof. (i) It follows from Observation 1.1(a) that k(|V−1| + |V1| + |V2| + |V3|) = kn

≤ 

v∈V (G)

f [v]

= 

v∈V (G)

(d(v) + 1)f (v)

= 

v∈V3

3(d(v) + 1) + 

v∈V2

2(d(v) + 1) + 

v∈V1

(d(v) + 1)

− 

v∈V−1

(d(v) + 1)

≤ 3(Δ+1)|V₃|+2(Δ+1)|V₂|+(Δ+1)|V₁| − (δ+1)|V₋₁|.

The inequality chain leads to the desired bound in (i).

(ii) By Observation 1.1(a), we have|V−1| = n − |V1| − |V2| − |V3|. By this identity and Part (i) of Proposition 2.1, we reach (ii).

(iii) According to Observation 1.1 and Part (ii) of Proposition 2.1, we obtain Part (iii) of Proposition 2.1 as follows.

(Δ + δ + 2)ω(f ) = (Δ + δ + 2)(2|V₁| + 3|V₂| + 4|V₃| − n)

≥ 2(δ + k + 1)n − 2(3Δ + δ + 4)|V3| − 2(2Δ + δ + 3)|V2| +(Δ + δ + 2)(3|V2| + 4|V3| − n)

= (δ− Δ + 2k)n + (δ − Δ)|V2| + 2(δ − Δ)|V3|.

(iv) The inequality chain in the proof of Part (i) and Observation 1.1(a) show that

kn ≤ 3(Δ + 1)|V3| + 2(Δ + 1)|V2| + (Δ + 1)|V1| − (δ + 1)|V−1|

≤ 3(Δ + 1)|V₁∪ V₂∪ V₃| − (δ + 1)|V₋₁|

= 3(Δ + 1)|V₁∪ V₂∪ V₃| − (δ + 1)(n − |V₁∪ V₂ ∪ V₃|)

= (3Δ + δ + 4)|V1∪ V2∪ V3| − (δ + 1)n and so

|V₁∪ V₂ ∪ V₃| ≥ n(δ + k + 1) 3Δ + δ + 4 . Using this equality and Observation 1.1, we obtain

ω(f ) = 2|V₁∪ V₂ ∪ V₃| − n + |V₂| + 2|V₃|

≥ n(δ− 3Δ + 2k − 2)

3Δ + δ + 4 +|V₂| + 2|V₃|.

This is the bound in Part (iv), and the proof is complete.

Proposition 2.2. Let r be a non-negative integer with r ≥ ^k−3₃ . If G is an r- regular graph of order n, then

γ_sdR^k (G)≥ kn r + 1.

The equality holds for the complete graph K_n when n + k≥ 6.

Proof. Let f = (V₋₁, V₁, V₂, V₃) be a γ^k_sdR(G)-function. We have (r + 1)γ_sdR^k (G) = (r + 1) 

v∈V (G)

f (v) = 

v∈V (G)

(r + 1)f (v) = 

v∈V (G)

f [v]≥ kn

and this leads to the desired bound. By Observation 1.3, the equality holds for the complete graph K_n if n + k ≥ 6.

Corollary 2.1. If G is a graph of order n, minimum degree δ≥ k −1 and maximum degree Δ, then

γ_sdR^k (G)≥−3Δ²+ 3Δδ + 4kΔ− 3Δ + 3δ + 4k (Δ + 1)(3Δ + δ + 4)

n.

Proof. If δ < Δ, multiplying both sides of the inequality in Proposition 2.1 (iv) by Δ− δ and adding the resulting inequality to the inequality in Proposition 2.1 (iii), we obtain the desired lower bound.

If δ = Δ = r, the desired inequality can be simplified to γ_sdR^k (G) ≥ _r+1^kn . It obviously holds according to Proposition 2.2.

Proposition 2.3. If G is a graph of order n with δ(G)≥ ^k−3₃ , then γ_sdR^k (G)≥ Δ(G) + k + 1 − n.

Proof. Let u∈ V (G) be a vertex of maximum degree, and let f be a γ^k_sdR(G)-function.

By the definitions we have γ_sdR^k (G) = f [u] + 

x∈V (G)−N [u]

f (x)≥ k − (n − Δ(G) − 1) = Δ(G) + k + 1 − n,

and the proof is complete.

A set S ⊆ V (G) is a 2-packing of the graph G if N[u] ∩ N[v] = ∅ for any two distinct vertices u, v∈ S. The 2-packing number ρ(G) of G is defined by

ρ(G) = max{|S| : S is a 2-packing of G}.

Clearly, for all graphs G, ρ(G)≤ γ(G).

Proposition 2.4. If G is a graph of order n with δ(G)≥ ^k−3₃ , then γ_sdR^k (G)≥ ρ(G)(δ(G) + k + 1) − n.

Proof. Let {v₁, v₂, . . . , v_ρ(G)} be a 2-packing of G, and let f be a γ^k_sdR(G)-function.

Suppose A =_ρ(G)

i=1 N [v_i]. We have |A| =^ρ(G)

i=1

(d(v_i) + 1)≥ ρ(G)(δ(G) + 1) and hence

γ_sdR^k (G) =

ρ(G)

i=1

f [v_i] + 

x∈V (G)−A

f (x)

≥ kρ(G) + 

x∈V (G)−A

f (x)≥ kρ(G) − n + |A|

≥ kρ(G) − n + ρ(G)(δ(G) + 1) = ρ(G)(δ(G) + k + 1) − n.

Corollary 2.2. If G is a graph of order n with δ(G) ≥ ^k−1₂ and ρ(G) = γ(G), then

γ_sdR^k (G)≥ (δ(G) + k + 1)γ(G) − n.

If G is the graph obtained from a graph H by adding a pendant edge at each vertex of H, then clearly γ_sdR^k (G) = 3n(G)/2 and hence the bound in Corollary 2.2 is sharp.

Since for any connected graph G, we have ρ(G) ≥ 1 + ^diam(G)₃ , the next result is an immediate consequence of Proposition 2.4.

Corollary 2.3. If G is a graph of order n with δ(G)≥ ^k−1₂ , then γ_sdR^k (G)≥

1 +diam(G) 3

(δ(G) + k + 1)− n.

Next we present a so-called Nordhous-Gaddum type inequality for the signed double Roman k-domination number of regular graphs.

Theorem 2.1. If G is an r-regular graph of order n such that r ≥ ^k−3₃  and n− r − 1 ≥ ^k−3₃ , then

γ_sdR^k (G) + γ_sdR^k (G)≥ 4kn n + 1. If n is even, then γ_sdR^k (G) + γ_sdR^k (G)≥ ^4k(n+1)_n+2 .

Proof. Since G is an r-regular, the complement G is (n− r − 1)-regular. By Propo- sition 2.2

γ_sdR^k (G) + γ_sdR^k (G)≥ kn 1

r + 1+ 1 n− r

.

By assumptions ^k−3₃  ≤ r ≤ n − 1 − ^k−3₃  and since the function g(x) = 1/(x + 1) + 1/(n− x) takes its minimum at x = (n − 1)/2 when ^k−3₃  ≤ x ≤ n − 1 − ^k−3₃ , we obtain

γ_sdR^k (G) + γ_sdR^k (G)≥ kn 2

n + 1+ 2 n + 1

 = 4kn n + 1,

and this is the desired bound. If n is even, then the function g takes its minimum at r = x = (n− 2)/2 or r = x = n/2, since r is an integer. This implies that

γ_sdR^k (G) + γ_sdR^k (G)≥ kn 1

r + 1 + 1 n− r

≥ kn 2

n + 2 n + 2

= 4k(n + 1) n + 2 , and the proof is complete.

Next we establish bounds on the signed double Roman k-domination number in terms of order and domination number.

Proposition 2.5. Let G be a connected graph of order n. If G has a γ_sdR^k (G)-function f = (V₋₁, V₁, V₂, V₃) with V₋₁ =∅, then γ_sdR^k (G)≥ n + γ(G).

Proof. Let f = (V₋₁, V₁, V₂, V₃) be a γ_sdR^k (G)-function such that V₋₁ =∅. Since each vertex in V₁ must be adjacent to a vertex in V₂ ∪ V₃, we deduce that V₂ ∪ V₃ is a dominating set of G. It follows from Observation 1.1 that

γ_sdR^k (G) = ω(f ) =|V₁| + 2|V₂| + 3|V₃| ≥ |V₁| + 2|V₂| + 2|V₃| ≥ n + γ(G).

Proposition 2.6. Let G be a graph of order n with δ(G)≥ 1. If k ∈ {2, 3}, then γ_sdR^k (G)≤ n + γ(G).

Furthermore, this bound is sharp.

Proof. Let S be a γ(G)-set and define f : V (G) → {−1, 1, 2, 3} by f(x) = 2 for x ∈ S and f(x) = 1 for x ∈ V (G) − S. Clearly, f is an SDRkDF on G of weight n + γ(G) for k∈ {2, 3} and this implies that γ_sdR^k (G)≤ n + γ(G) for k ∈ {2, 3}.

To prove the sharpness, let G₂ = mK₁ and let G₃ be the graph obtained from a graph H by adding a pendant edge at each vertex of H. Clearly, for any γ_sdR^k (G_k)- function f = (V₋₁, V₁, V₂, V₃) we have V₋₁ =∅. It follows from Proposition 2.5 that γ_sdR^k (G_k)≥ n(Gk) + γ(G_k) yielding γ_sdR^k (G_k) = n(G_k) + γ(G_k).

The proof of next result is similar to the proof of Proposition 2.6 and therefore it is omitted.

Proposition 2.7. For any graph G of order n with δ(G)≥ 1, γ_sdR⁴ (G)≤ n + 2γ(G).

The bound is sharp for the graph obtained from a graph H by adding at least two pendant edges at each vertex of H.

3 Signed double Roman 2-domination

In this section, we present bounds on the signed double Roman 2-domination of G.

For convenience, we introduce some notation. For an SDR2DF f = (V₋₁, V₁, V₂, V₃) of G, we let V₋₁^ = {v ∈ V−1 | N(v) ∩ V3 = ∅} and V₋₁^ = V₋₁ − V₋₁^ . For a subset S ⊆ V , we let dS(v) denote the number of vertices in S that are adjacent to v. In particular, d_V(v) = d(v). For disjoint subsets U and W of vertices, we let [U, W ] denote the set of edges between U and W . For notational convenience, we let V₁₂ = V₁∪V₂, V₁₃= V₁∪V₃, V₁₂₃ = V₁∪V₂∪V₃ and let|V₁₂| = n₁₂,|V₁₃| = n₁₃,|V₁₂₃| = n₁₂₃, and let |V1| = n1,|V2| = n2 and |V3| = n3. Then, n₁₂₃ = n₁ + n₂ + n₃. Further, we let |V−1| = n−1, and so n₋₁ = n− n123. Let G₁₂₃ = G[V₁₂₃] be the subgraph induced by the set V₁₂₃ and let G₁₂₃ have size m₁₂₃. For i = 1, 2, 3, if V_i = ∅, let G_i = G[V_i] be the subgraph induced by the set V_i and let G_i have size m_i. Hence, m₁₂₃ = m₁+ m₂+ m₃+|[V1, V₂]| + |[V1, V₃]| + |[V2, V₃]|.

For t ≥ 1, let L_t be the graph obtained from a graph H of order t by adding 3d_H(v) + 1 pendant edges to each vertex v of H. LetH = {Lt| t ≥ 1}.

Theorem 3.1. Let G be a graph of order n and size m without isolated vertex. Then γ_sdR² (G)≥ 5n− 6m

2 ,

with equality if and only if G∈ H.

Proof. Let f = (V₋₁, V₁, V₂, V₃) be a γ_sdR² (G)-function. If V₋₁ = ∅, then γ_sdR² (G) >

n≥ ^5n−6m₂ since G has no isolated vertex. Hence V₋₁ = ∅. We consider the following cases.

Case 1. V₃ = ∅.

Now, we consider the following subcases.

Subcase 1.1. V₂ = ∅.

By the definition of an SDR2DF, each vertex in V₋₁ is adjacent to at least one vertex in V₃ or to at least two vertices in V₂, and so

|[V−1, V₁₂₃]| ≥ |[V−1, V₃]| + |[V−1, V₂]| ≥ |V₋₁^ | + 2|V₋₁^ | ≥ n−1. Furthermore we have

2n₋₁ ≤ 2|[V₋₁, V₃]| + |[V₋₁, V₂]| = 2

v∈V3

d_V−1(v) + 

v∈V2

d_V−1(v).

For each vertex v∈ V3, we have that f (v) + 3d_V3(v) + 2d_V2(v) + d_V1(v)− dV−1(v) = f [v]≥ 2, and so d_V−1(v)≤ 3d_V3(v) + 2d_V2(v) + d_V1(v) + 1. Similarly, for each vertex v ∈ V2, we have that d_V−1(v)≤ 3dV3(v) + 2d_V2(v) + d_V1(v). Now, we have

2n₋₁ ≤ 2

v∈V3

d_V−1(v) + 

v∈V2

d_V−1(v)

≤ 2

v∈V3

(3d_V3(v) + 2d_V2(v) + d_V1(v) + 1) +

v∈V2

(3d_V3(v) + 2d_V2(v) + d_V1(v))

= (12m₃+ 4|[V2, V₃]| + 2|[V1, V₃]| + 2n3) + (3|[V2, V₃]| + 4m2+|[V1, V₂]|)

= 12m₃+ 4m₂+ 7|[V2, V₃]| + 2|[V1, V₃]| + |[V1, V₂]| + 2n3

= 12m₁₂₃− 12m₁− 8m₂− 5|[V₂, V₃]| − 10|[V₁, V₃]| − 11|[V₁, V₂]| + 2n₃,

which implies that m₁₂₃ ≥ 1

12(2n₋₁+ 12m₁+ 8m₂+ 5|[V2, V₃]| + 10|[V1, V₃]| + 11|[V1, V₂]| − 2n3).

Hence,

m≥ m₁₂₃+|[V₋₁, V₁₂₃]| + m₋₁

≥ m123+|[V−1, V₁₂₃]|

≥ 1

12(2n₋₁+ 12m₁+ 8m₂+ 5|[V2, V₃]| + 10|[V1, V₃]| + 11|[V₁, V₂]| − 2n₃) + n₋₁

= 1

12(14n₋₁− 2n₁₂₃+ 2n₁+ 2n₂

+ 12m₁+ 8m₂+ 5|[V2, V₃]| + 10|[V1, V₃]| + 11|[V1, V₂]|)

= 1

12(14n− 16n123+ 2n₁+ 2n₂

+ 12m₁+ 8m₂+ 5|[V2, V₃]| + 10|[V1, V₃]| + 11|[V1, V₂]|)

and so n₁₂₃ ≥ 1

16(−12m+14n+2n₁+2n₂+12m₁+8m₂+5|[V₂, V₃]|+10|[V₁, V₃]|+11|[V₁, V₂]|).

Now, we have

γ_sdR² (G) = 3n₃+ 2n₂+ n₁− n₋₁

= 4n₃+ 3n₂+ 2n₁ − n

= 4n₁₂₃− n − n2− 2n1

≥ 1

4(−12m + 14n + 2n1+ 2n₂+ 12m₁+ 8m₂

+ 5|[V₂, V₃]| + 10|[V₁, V₃]| + 11|[V₁, V₂]|) − n − n₂− 2n₁

= 1

4(−12m + 14n − 4n) +1

4(2n₁+ 2n₂+ 12m₁+ 8m₂ + 5|[V2, V₃]| + 10|[V1, V₃]| + 11|[V1, V₂]| − 4n2− 8n1)

= 5n− 6m

2 + 1

4(−6n1− 2n2 + 12m₁+ 8m₂ + 5|[V2, V₃]| + 10|[V1, V₃]| + 11|[V1, V₂]|).

Let Θ =−6n1− 2n2+ 12m₁+ 8m₂+ 5|[V2, V₃]| + 10|[V1, V₃]| + 11|[V1, V₂]|. We show that Θ≥ 0. If n1 = 0, then Θ =−2n2+ 8m₂+ 5|[V2, V₃]|. If v ∈ V2 and d_V23(v) = 0, then since G has no isolated vertex, we have that every neighbor of v belongs to V₋₁. But then f [v]≤ 1, a contradiction. Hence if v ∈ V2, then we have that d_V23(v)≥ 1.

Then

Θ =−2n₂ + 8m₂ + 5|[V₂, V₃]|

= 4

v∈V2

d_V2(v) + 4

v∈V2

d_V3(v) + (|[V2, V₃]| − 2n2)

= 4

v∈V2

d_V23(v) + (|[V2, V₃]| − 2n2)

≥ 4n2− 2n2+|[V2, V₃]|

> 0.

If n₁ ≥ 1, then Θ = −6n₁− 2n₂+ 12m₁+ 8m₂+ 5|[V₂, V₃]| + 10|[V₁, V₃]| + 11|[V₁, V₂]|.

By the definition of an SDR2DF of G we have d_V123(v)≥ 1 for each v ∈ V1. Then Θ = −6n1− 2n2+ 12m₁+ 8m₂+ 5|[V2, V₃]| + 10|[V1, V₃]| + 11|[V1, V₂]|

= 6

v∈V1

d_V1(v) + 6

v∈V1

d_V2(v) + 6

v∈V1

d_V3(v) + 4

v∈V2

d_V1(v) + 4

v∈V2

d_V2(v) + 4

v∈V2

d_V3(v) + (−6n₁− 2n₂+|[V₂, V₃]| + 4|[V₁, V₃]| + |[V₁, V₂]|)

= 6

v∈V1

d_V123(v) + 4

v∈V2

d_V123(v) + (−6n1− 2n2+|[V2, V₃]| + 4|[V1, V₃]| + |[V1, V₂]|)

≥ 6n1+ 4n₂− 6n1 − 2n2+|[V2, V₃]| + 4|[V1, V₃]| + |[V1, V₂]|

= 2n₂+|[V₂, V₃]| + 3|[V₁, V₃]|

> 0.

Therefore γ²_sdR(G) > ^5n−6m

2 . Subcase 1.2. V₂ =∅.

By the definition of an SDR2DF, each vertex in V₋₁ is adjacent to one vertex in V₃, and so

|[V−1, V₁₃]| ≥ |[V−1, V₃]| ≥ |V−1| = n−1. (1) Furthermore, we have

n₋₁ ≤ |[V₋₁, V₃]| = 

v∈V3

d_V−1(v).

For each vertex v ∈ V3, we have that f (v) + 3d_V3(v) + d_V1(v)− dV−1(v) = f [v] ≥ 2, and so d_V−1(v)≤ 3d_V3(v) + d_V1(v) + 1. Now, we have

n₋₁ ≤ 

v∈V3

d_V−1(v) (2)

≤ 

v∈V3

(3d_V3(v) + d_V1(v) + 1)

= 6m₃+|[V1, V₃]| + n3

= 6m₁₃− 6m₁ − 5|[V₁, V₃]| + n₃, which implies that

m₁₃ ≥ 1

6(n₋₁+ 6m₁+ 5|[V₁, V₂]| − n₃).

Hence,

m ≥ m13+|[V−1, V₁₃]| + m−1

≥ m13+|[V−1, V₁₃]|

≥ 1

6(n₋₁+ 6m₁+ 5|[V₁, V₂]| − n₃) + n₋₁

= 1

6(7n₋₁− n₁₃+ n₁+ 6m₁+ 5|[V₁, V₃]|)

= 1

6(7n− 8n₁₃+ n₁+ 6m₁+ 5|[V₁, V₃]|)

and so

n₁₃ ≥ 1

8(−6m + 7n + n₁+ 6m₁+ 5|[V₁, V₃]|).

Now, we have

γ_sdR² (G) = 3n₃+ n₁− n₋₁

= 4n₃+ 2n₁ − n

= 4n₁₃− n − 2n1

≥ 1

2(−6m + 7n + n1+ 6m₁ + 5|[V1, V₃]|) − n − 2n1 (3)

= 1

2(−6m + 7n − 2n) + 1

2(n₁ + 6m₁ + 5|[V1, V₃]| − 4n1)

= 5n− 6m

2 +1

2(6m₁+ 5|[V1, V₃]| − 3n1).

Let Θ = 6m₁ + 5|[V1, V₃]| − 3n1. We show that Θ ≥ 0. If n1 = 0, then Θ = 0.

Suppose that n₁ ≥ 1. By the definition of an SDR2DF, for each v ∈ V₁ we have d_V13(v)≥ 1. Then

Θ = 6m₁+ 5|[V₁, V₃]| − 3n₁

= 3

v∈V1

d_V1(v) + 3

v∈V1

d_V3(v) + (2|[V1, V₃]| − 3n1)

= 3

v∈V1

d_V13(v) + (2|[V1, V₃]| − 3n1)

≥ 3n₁ + 2|[V₁, V₃]| − 3n₁

> 0.

Therefore γ²_sdR(G)≥ ^5n−6m₂ . Case 2. V₃ =∅.

Since V₋₁ = ∅, we conclude that V2 = ∅. By the definition of an SDR2DF, each vertex in V₋₁ is adjacent to at least two vertices in V₂, and so

|[V−1, V₁₂]| ≥ |[V−1, V₂]| ≥ 2|V−1| = 2n−1. Furthermore we have

2n₋₁ ≤ |[V−1, V₂]| = 

v∈V2

d_V−1(v).

For each vertex v ∈ V2, we have that f (v) + 2d_V2(v) + d_V1(v)− dV−1(v) = f [v] ≥ 2, and so d_V−1(v)≤ 2dV2(v) + d_V1(v). Now, we have

2n₋₁ ≤ 

v∈V2

d_V−1(v)

≤ 

v∈V2

(2d_V2(v) + d_V1(v))

= 4m₂+|[V1, V₂]|

= 4m₁₂− 4m₁ − 3|[V₁, V₂]|,

which implies that

m₁₂≥ 1

4(2n₋₁+ 4m₁ + 3|[V1, V₂]|).

Hence,

m≥ m12+|[V−1, V₁₂]| + m−1

≥ m12+|[V−1, V₁₂]|

≥ 1

4(2n₋₁+ 4m₁+ 3|[V1, V₂]|) + 2n−1

= 1

4(10n₋₁+ 4m₁+ 3|[V1, V₂]|)

= 1

4(10n− 10n12+ 4m₁+ 3|[V1, V₂]|)

and so

n₁₂ ≥ 1

10(−4m + 10n + 4m1+ 3|[V1, V₂]|).

Now, we have

γ_sdR² (G) = 2n₂+ n₁− n₋₁

= 3n₂+ 2n₁− n

= 3n₁₂− n − n₁

≥ 3

10(−4m + 10n + 4m1+ 3|[V1, V₂]|) − n − n1

= 3

10(−4m + 10n −10 3 n− 1

3n) + 3

10(4m₁+ 3|[V₁, V₂]| − 10

3 n₁ +1 3n)

= 19n− 12m

10 + 3

10(4m₁+ 3|[V₁, V₂]| − 3n₁+1

3(n₋₁+ n₂))

Let Θ = 4m₁+ 3|[V1, V₂]| − 3n1+¹

3(n₋₁+ n₂). We show that Θ > 0. If n₁ = 0, then Θ = ¹₃(n₂+ n₋₁) > 0. Suppose that n₁ ≥ 1. By the definition of an SDR2DF of G, we have d_V2(v)≥ 1 for each v ∈ V₁. Then

Θ = 4m₁ + 3|[V₁, V₂]| − 3n₁ +1

3(n₋₁+ n₂)

= 3

v∈V1

d_V2(v) + (1

3(n₂+ n₋₁)− 3n₁+ 4m₁)

≥ 3n₁+1

3(n₂+ n₋₁)− 3n₁+ 4m₁

> 0 Therefore γ²_sdR(G) > ^19n−12m

10 . Since G has no isolated vertex, we have γ_sdR² (G) > 19n− 12m

10 > 5n− 6m

2 .

Let γ_sdR² (G) = ^5n−6m

2 . Then all inequalities (1), (2) and (3) must be equalities.

In particular, n₁ = 0 and n₃ = n₁₃, and so V₁₃= V₃ and V = V₃∪ V−1. Furthermore, m = m₃+|[V−1, V₃]|, |[V−1, V₃]| = n−1 and m₃ = ¹₆(n₋₁− n3). This implies that for each vertex v ∈ V−1 we have d_V−1(v) = 0 and d_V3(v) = 1, and so each vertex of V₋₁ is a leaf in G. Moreover for each vertex v ∈ V₃ we have d_V−1(v) = 3d_V3(v) + 1. Hence G∈ H.

On the other hand, let G∈ H. Then G = Lt for some t≥ 1. Thus, G is obtained from a graph H of order t by adding 3d_H(v) + 1 pendant edges to each vertex v of H. Let G have order n and size m. Then,

n = 

v∈V (H)

(3d_H(v) + 2) = 6m(H) + 2n(H)

and

m = m(H) + 

v∈V (H)

(3d_H(v) + 1) = 7m(H) + n(H).

Assigning to every vertex in V (H) the weight 3 and to every vertex in V (G)− V (H) the weight−1 produces an SDR2DF f of weight ω(f) = 3n(H)−(6m(H)+n(H)) = 2n(H)− 6m(H) = ^5n−6m₂ . Hence γ_sdR² (G)≤ ^5n−6m₂ . It follows that γ_sdR² (G) = ^5n−6m₂ and this completes the proof.

Theorem 3.2. Let G be a graph of order n≥ 3. Then γ²_sdR(G)≥ 4

n + 2 3 − n.

This bound is sharp for DS_4,4.

Proof. Let f = (V₋₁, V₁, V₂, V₃) be a γ_sdR² (G)-function. If |V−1| = 0, then γ_sdR² (G) >

n≥ 4

n+2

3 − n. Hence V−1 = ∅. We consider the following cases.

Case 1. V₃ = ∅.

Since each vertex in V₋₁^ is adjacent to at least one vertex in V₃, we conclude that at least one vertex v of V₃ is adjacent to at least ⁿ_n^−1

3 vertices of V₋₁^ . Also, since each vertex in V₋₁^ is adjacent to at least two vertices in V₂, we conclude that at least one vertex u of V₂ is adjacent to at least ^2n−1

n2 vertices of V₋₁^ . Then 2≤ f[v] ≤ 3n₃+ 2n₂+ n₁− ⁿ_n^−1₃ which implies that

0≤ 3n²₃+ 2n₂n₃+ n₁n₃− n^₋₁− 2n3. (4) Similarly, we have

0≤ 3n3n₂+ 2n²₂+ n₁n₂− 2n^₋₁− 2n2. (5) By multiplying the inequality (4) by 2 and summing it with the inequality (5), we obtain

0≤ 6n²₃+ 2n²₂+ 7n₂n₃+ 2n₁n₃+ n₁n₂− 2n^₋₁− 2n^₋₁− 2n2− 4n3.

Since n = n₃+ n₂+ n₁+ n₋₁, we have

0≤ 6n²₃ + 2n²₂+ 7n₂n₃+ 2n₁n₃+ n₁n₂ + 2n₁− 2n3− 2n, equivalently

0≤ 16n²₃+16

3 n²₂+56

3 n₂n₃+16

3 n₁n₃+16

6 n₁n₂+16

3 n₁− 16

3 (n + n₃)

≤ 16n²₃+ 9n²₂+ 4n²₁ + 24n₂n₃+ 16n₁n₃+ 12n₁n₂−16

3 (n + n₃)

= (4n₃+ 3n₂+ 2n₁)²− 16

3 (n + n₃) which implies that 4

n+n3

3 ≤ 4n3+ 3n₂+ 2n₁. If n₃ ≥ 2, then 4

n+2

3 ≤ 4n3+ 3n₂+ 2n₁. Hence let n₃ = 1. Then

0≤ 16 + 16

3 n²₂+ 56

3 n₂ +32

3 n₁+16

6 n₁n₂− 16

3 (n + 1)

≤ 16 + 9n²₂+ 4n²₁+ 24n₂ + 16n₁+ 12n₁n₂ −16

3 (n + 1)− 16

3 (n₁+ n₂)

= (4n₃+ 3n₂+ 2n₁)²− 16

3 (n + n₁ + n₂+ 1) which implies that 4

n+1+n1+n2

3 ≤ 4n₃+ 3n₂ + 2n₁. Since n≥ 3, we conclude that n₁+ n₂ ≥ 1, and so

4

n + 2

3 ≤ 4n3 + 3n₂+ 2n₁. Therefore

γ_sdR(G) = 3n₃+ 2n₂ + n₁− n−1

= 4n₃+ 3n₂ + 2n₁− n

≥ 4

n + 2 3 − n.

Case 2. V₃ =∅.

Since V₋₁ = ∅, we conclude that V₂ = ∅. As in Case 1, at least one vertex u of V₂ is adjacent to at least ²ⁿ⁻¹

n2 vertices of V₋₁. Then 2 ≤ f[u] ≤ 2n2+ n₁− ²ⁿ_n⁻¹₂ which implies that

0≤ 2n²₂+ n₁n₂− 2n₋₁− 2n₂. Since n = n₂+ n₁ + n₋₁, we have

0≤ 2n²₂+ n₁n₂+ 2n₁− 2n, equivalently

0≤ 16

3n²₂+16

6 n₁n₂+16

3 n₁− 16 3 n

≤ 9n²₂ + 4n²₁+ 12n₁n₂− 16

3 n− 3n²₂

= (3n₂+ 2n₁)²− 16

3 n− 3n²₂.